Timeline for On the group actions on Hurwitz surfaces
Current License: CC BY-SA 3.0
7 events
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| Feb 17, 2013 at 20:16 | comment | added | Peter Mueller | @Klim: This should be in any book on Riemann surfaces. The proof goes as follows: Let $b_1,b_2,\dots,b_r$ be the branch points of the cover $C\to P^1$. Let $b\in P^1$ be different from the branch points. Then the fundamental group of $P^1$ minus the branch points, with base point $b$, is generated by loops around $b_i$, and if chosen properly, these generators' product is $1$. The fundamental group acts on the fiber over $b$ by lifting of paths, and the induced permutation group is isomorphic to the group of deck transformations, which in this case is $G$. | |
| Feb 17, 2013 at 14:34 | comment | added | Klim Puhov | @Peter: Ok, maybe "classification" is not a right word. So, you use the description of finite branched Galois covers of $P^1$. Where is the proof can be found? | |
| Feb 17, 2013 at 14:02 | comment | added | Peter Mueller | @Klim: I do not agree that I use any classification theorems. If $G$ is the automorphism group of a Hurwitz surface $C$, then $C/G$ is a projective line $P^1$ (should be a by-product in the proof of the Hurwitz bound). Finite branched Galois covers with group $G$ of $P^1$ are described in terms of generating systems $g_1,\dots,g_r$ of $G$ with $g_1g_2\dots g_r=1$, where the $g_i$ are the generators of the local monodromies. Knowing that in the Hurwitz case there are three generators of orders $2$, $3$, and $7$ respectively is more or less rephrasing that the Hurwitz bound is sharp. | |
| Feb 17, 2013 at 11:18 | comment | added | Klim Puhov | @Peter: Your answer is exactly what I called "using of classification theorems". So, unfortunately, it is not helpfull for me. Using Riemann-Hurwitz genus formula, together with the proof of the Hurwitz bound, I can show that the quotient map by the $G$-action have ramification points of indexes $2$, $3$ and $7$, but I can't figure out that $G$ have the description that you give in your answer. | |
| Feb 17, 2013 at 10:39 | comment | added | Peter Mueller | Well, many people even take that as the definition, as in the paper you quoted in your comment. I believe that the Riemann-Hurwitz genus formula, together with the proof of the Hurwitz bound, gives this connection. | |
| Feb 17, 2013 at 10:26 | comment | added | Klim Puhov | Could you explain, please, why $G$ is generated by a,b,c with relatively prime orders 2,3,7 and abc=1? I can't find the proof in literature. | |
| Feb 17, 2013 at 10:15 | history | answered | Peter Mueller | CC BY-SA 3.0 |